If I had a mass of $100\:\rm{kg}$ accelerating due to gravity, using $F=ma$:
$F = 100\:\rm{kg} \times 9.8\:\rm{m/s^2}$
$F = 980 \:\rm N$...
If I increased the mass to 200kg, the force would be 1960 N:
$F = 200\:\rm{kg} \times 9.8\:\rm{m/s^2}$
$F = 1960 \:\rm{N}$
Now, finally getting to my question: Does this increase in force (which is supposed to be a push/pull) mean that the object would fall faster when it weighs more?
No, the heavier object does not fall faster. Instead, they heavy and light object fall at the same acceleration (and hence the same speed if they are both simply dropped). This is an example of the equivalence principle.
The more massive object has more gravitational force on it, but it also has more inertia. Specifically, because the object is twice as massive, it has twice the inertial mass.
The force on it is doubled, so the acceleration stays the same.
If we look at
$$F = ma$$
we see that when $F$ and $m$ are both multiplied by 2, $a$ stays the same.
Check these questions for more:
Free falling of object with no air resistance
Agreed that the answer to the question is, "No." Acceleration remains constant.
One way to think of it is this: The first 980N is accelerating the first 100kg at 9.8m/(s^2). The second 980N is accelerating the second 100kg at 9.8m/(s^2). Both masses fall at the same velocity and acceleration (neglecting wind resistance, etc.). So it should be easy to see how you could join these to 100kg masses into one 200kg mass, with 1960N pulling it down, without changing the speed or acceleration of either mass.
Lets assume there are no non-conservative forces like air drag and you are dropping the ball from rest. While the ball gains kinetic energy, it loses potential energy. This means that
$$ K_\text{in} + P_\text{in} = K_\text{f} + P_\text{f} $$ $$ 0 + mgh = \frac{1}{2} mv^2 + 0 $$
which means $v= \sqrt{2gh}$ and there is no mass term here. Hence the (final) velocity is not dependent on weight.
